A few areas of algebra I don't understand

As I mentioned in another post I'm trying to keep up with my math class and since I didn't finish high school theres some things I have to learn on my own. Heres a few things that I didn't figure out in the class. This is in chronological order, things are gradually getting more complex so I'd better keep up now if I'm gonna have any chance of doing well when it gets hard. We're still in the warmup period.

1.) Unsquaring short equations
Q.) (2A + B)sq A.) 4Asq + Bsq + 4ab
I would have just answered it 4Asq + Bsq I have no idea where the 4ab comes from.

2.) Turning equations with x to the power of 3 into quadratic equations
I'm fairly lost here. Heres an example question.
Q.) Xcubed + 5Xsq - 12X - 16 divided by X + 1
I know how to do the division but I'm wondering what its all about. The aim seems to be to turn this cubed equation into a simpler quadratic equation. To do this can I just divide any of its factors into it?

3.) Even harder cubed equations. The teacher started throwing the coefficient "K" into the equations. Example:
Q.) 2Xcubed - Xsq + KX - 12 divided by 2X + 1
I have no idea what to do with this K

I'm having trouble keeping up with the class because when I take time to go over one thing he moves onto another and I gradually fall behind more and more so I have to figure out as much of my notes as I can when I go home. One problem is I don't know the language of math so I have trouble searching for help on certain things I'm having trouble with. I'm glad theres forums like this around.

Now, an algebraic expression with two terms is called a binomial. The question is, how do we multiply two binomials together? The answer comes from the fact that multiplication is distributive: *each* term in the first binomial has to be multiplied by every term in the second binomial (i.e. it gets distributed through), and vice versa. So the expression becomes:

Note that the two middle terms (which are called the "cross" terms because they involve multiplying the *unlike* terms from each binomial) are identical and can be added together to give you 2ab + 2ab = 4ab. This is true generally and gives you a bit of a shortcut for squaring a binomial. Anyway, so our result is:

3.) Even harder cubed equations. The teacher started throwing the coefficient "K" into the equations. Example:
Q.) 2Xcubed - Xsq + KX - 12 divided by 2X + 1
I have no idea what to do with this K

k is just a constant, unlike x, which is a variable. As the name suggests, the value of a variable can change, whereas you can think of a constant has having some fixed value. So having a k there is no different than having a "2" or a "3" or any other number for that matter. He's just simply not specifying what the value of k is. It's some constant.

Very simple example:

I have 2x + kx. How can I "perform addition" in order to rewrite this? Well, the first term says that I have 2 times x, i.e. I have two copies of "x". The second terms says that I have "k" copies of "x", whatever k happens to be. So how many copies of x do I have in total? Answer: Just add together the copies to get a total number of (k + 2) copies

So we can rewrite 2x + kx as (2 + k)x or as (k + 2)x. (The order in which addition is performed doesn't matter. We say that addition is "commutative").

This is an example of "collecting" *like* terms. Both terms in this example are terms involving x multiplied by some coefficient. Since both terms contain x, they can be collected together, which amounts to keep track of the total number of "x's" you have in your equation.

Q.) Xcubed + 5Xsq - 12X - 16 divided by X + 1
I know how to do the division but I'm wondering what its all about. The aim seems to be to turn this cubed equation into a simpler quadratic equation. To do this can I just divide any of its factors into it?

I don't know what you mean. The aim is not to magically turn the cubic expression into a quadratic one. The aim is to perform the long division i.e. to figure out an algebraic expression for the quotient of these two polynomials -- this expression will truly be their quotient no matter what value x has.

Since you are dividing a cubic by a linear polynomial, the result will be quadratic in *this* case.

As I mentioned in another post I'm trying to keep up with my math class and since I didn't finish high school theres some things I have to learn on my own. Heres a few things that I didn't figure out in the class. This is in chronological order, things are gradually getting more complex so I'd better keep up now if I'm gonna have any chance of doing well when it gets hard. We're still in the warmup period.

Just keep at it! Ask for additional help from here and your teacher where you need it.

1.) Unsquaring short equations
Q.) (2A + B)sq A.) 4Asq + Bsq + 4ab
I would have just answered it 4Asq + Bsq I have no idea where the 4ab comes from.

Be careful about your terminology. We do not call this "unsquaring".

In algebra, we have a number of rules associated with arithmetic operations. Here is a quick search result on Google listing some of the rules:

The poster above me explained roughly how foil works. FOIL is your best friend in algebra. Memorize it and learn to love it =-)

2.) Turning equations with x to the power of 3 into quadratic equations
I'm fairly lost here. Heres an example question.
Q.) Xcubed + 5Xsq - 12X - 16 divided by X + 1
I know how to do the division but I'm wondering what its all about. The aim seems to be to turn this cubed equation into a simpler quadratic equation. To do this can I just divide any of its factors into it?

Again, please be careful with terminology. We don't "turn cubic equations into quadratic equations". The process you are describing is called "factorization". We want to "factorize" [tex]x^3 + 5x^2 - 12x - 16[/tex].

You should be familiar with prime factorization: that for every integer, you can break it down into prime components.

6 = 2 * 3
12 = 2 * 2 * 3
121 = 11 * 11
17 = 17

Each set of numbers on the right hand side is a prime number. In certain cases (namely, for prime numbers such as 17), there is no way to break the number down further.

For problems two and three there is a hint in what you are dividing the cubic equation by to help you factor it.

x^3+5x^2-12x-16/x+1

x+1 can be factored out of the denominator

(x+1)(ax^2+bx+c)/(x+1)

x+1 cancels out and you are left with just the quadratic to solve.

Whenever you are dividing one ploynomial by another first try to factor the polynomial that is in the denominator out of the numerator, this will save you the hassle of having to try and divide two polynomials. Keep in mind it won't work out everytime like it did in this one, so don't always expect it. Just try it first everytime to see.